University Quantum Mechanics I · topic UQMI-10.07
Matrix representation of operators: A_mn = <m|A|n>
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Matrix representation of operators: Aₘₙ = ⟨m|Â|n⟩: predict a finite change
The matrix of an operator  in an orthonormal basis |n⟩ is the array of numbers Aₘₙ = ⟨m|Â|n⟩, whose column n holds the components of Â|n⟩; multiplying it into the column cₙ = ⟨n|ψ⟩ gives the components of Â|ψ⟩. Use the supplied topic model Y_out = (cosd(t)*cosd(t)*0.2826727+sind(t)*sind(t)*0.3206682-0.180127*sind(2*t)), where Y_out is ⟨x²⟩ FROM ALL ROWS (L²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. t is Mix θ in cos θ ψ₁ + sin θ ψ₂, measured in °. Keep n (Rows of xₘₙ kept, N) = 2 fixed. For this exercise use only 19.125 ° ≤ t ≤ 38.025 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at t = 23.283 ° and 25.929 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
⟨x²⟩ FROM ALL ROWS against t
Predictions from the supplied Matrix representation of operators: Aₘₙ = ⟨m|Â|n⟩ model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): ⟨x²⟩ FROM ALL ROWS / L²
Horizontal axis (x): t: Mix θ in cos θ ψ₁ + sin θ ψ₂ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Mix θ in cos θ ψ₁ + sin θ ψ₂ / ° | ⟨x²⟩ FROM ALL ROWS / L² |
|---|---|
| 19.125 | 0.17523565946902786 |
| 23.85 | 0.15565719627625765 |
| 28.575 | 0.1400417497137035 |
| 33.3 | 0.12881314620011464 |
| 38.025 | 0.12227614671659098 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 23.283 °, substitute into Y_out = (cosd(t)*cosd(t)*0.2826727+sind(t)*sind(t)*0.3206682-0.180127*sind(2*t)): Y₀ = 0.157807 L².
- At t = 25.929 °, the same fixed-parameter model gives Y₁ = 0.14827 L².
- Signed change: ΔY_out = Y₁ − Y₀ = -0.00953643 L²; the input change is Δt = 2.646 °.
- Average sensitivity is ΔY_out/Δt = -0.00360409 L² per °. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.157807 L², Y₁ = 0.14827 L²; ΔY_out = -0.00953643 L²; average sensitivity = -0.00360409 L² per °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Matrix representation of operators: Aₘₙ = ⟨m|Â|n⟩: infer an input with uncertainty
The matrix of an operator  in an orthonormal basis |n⟩ is the array of numbers Aₘₙ = ⟨m|Â|n⟩, whose column n holds the components of Â|n⟩; multiplying it into the column cₙ = ⟨n|ψ⟩ gives the components of Â|ψ⟩. Use the supplied topic model Y_out = (cosd(t)*cosd(t)*0.2826727+sind(t)*sind(t)*0.3206682-0.180127*sind(2*t)), where Y_out is ⟨x²⟩ FROM ALL ROWS (L²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. t is Mix θ in cos θ ψ₁ + sin θ ψ₂, measured in °. Keep n (Rows of xₘₙ kept, N) = 2 fixed. For this exercise use only 19.125 ° ≤ t ≤ 38.025 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.13453983 L², with a bounded reading uncertainty ±0.0018396 L². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00189 °, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
⟨x²⟩ FROM ALL ROWS against t
Predictions from the supplied Matrix representation of operators: Aₘₙ = ⟨m|Â|n⟩ model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): ⟨x²⟩ FROM ALL ROWS / L²
Horizontal axis (x): t: Mix θ in cos θ ψ₁ + sin θ ψ₂ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Mix θ in cos θ ψ₁ + sin θ ψ₂ / ° | ⟨x²⟩ FROM ALL ROWS / L² |
|---|---|
| 19.125 | 0.17523565946902786 |
| 23.85 | 0.15565719627625765 |
| 28.575 | 0.1400417497137035 |
| 33.3 | 0.12881314620011464 |
| 38.025 | 0.12227614671659098 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(19.125) = 0.175236 L² and Y_out(38.025) = 0.122276 L². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 19.125 + (0.13454 − 0.175236)(38.025 − 19.125)/(0.122276 − 0.175236) = 33.6484 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t = 30.654 °; substituting back gives Y_out = 0.13453983 L².
- The interpolation error in the input is |t_line − t| = 2.99438 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.13270023 L² and 0.13637943 L². Sort the roots even if the response decreases: 29.9215 ° ≤ t ≤ 31.4344 °.
- The lower and upper input deviations are 0.732532 ° and 0.780414 °. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ 30.654 °; compatible interval [29.9215, 31.4344] °. Endpoint interpolation gives 33.6484 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.